Question: Simplify the following expression and state the condition under which the simplification is valid: $y = \dfrac{r^2 + 3r}{r^2 - 9}$
Answer: First factor the expressions in the numerator and denominator. $ \dfrac{r^2 + 3r}{r^2 - 9} = \dfrac{(r)(r + 3)}{(r - 3)(r + 3)} $ Notice that the term $(r + 3)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(r + 3)$ gives: $y = \dfrac{r}{r - 3}$ Since we divided by $(r + 3)$, $r \neq -3$. $y = \dfrac{r}{r - 3}; \space r \neq -3$